WAEC Further Mathematics Past Questions & Answers - Page 54

Equation of a circle: $(x - a)^{2} + (y - b)^{2} = r^{2}$

Given that $x^{2} + y^{2} - 4x + 8y + 11 = 0$

Expanding the equation of a circle, we have: $x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}$

Comparing this expansion with the given equation, we have

$2a = 4 \implies a = 2$

$-2b = 8 \implies b = -4$

$r^{2} - a^{2} - b^{2} = -11 \implies r^{2} = -11 + 2^{2} + 4^{2} =9$

$r = 3$

$Area = \pi r^{2} = \pi \times 3^{2}$

= $9\pi$

$m . n = |m||n|\cos \theta$

$(3i + 4j) . (2i - j) = 6 - 4 = 2$

$2 = |(3i + 4j)||(2i - j)| \cos \theta$

$|3i + 4j| = \sqrt{3^{2} + 4^{2}} = \sqrt{25} = 5$

$|2i - j| = \sqrt{2^{2} + (-1)^{2}} = \sqrt{5}$

$2 = 5(\sqrt{5})(\cos \theta)$

$\cos \theta = \frac{2}{5\sqrt{5}} = 0.08\sqrt{5}$ 

$\theta = \cos^{-1} 0.1788 = 79.7°$

$\log_{2} y^{\frac{1}{2}} = \log_{5} 125$

$\log_{2} y^{\frac{1}{2}} = \log_{5} 5^{3} = 3\log_{5} 5 = 3$

$\log_{2} y^{\frac{1}{2}} = 3 \implies y^{\frac{1}{2}} = 2^{3} = 8$

$y = 8^{2} = 64$

$y = 4x^{3} + kx^{2} - 6x + 4$

$\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6$

At P(1, m)

$\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0$ (parallel to the x- axis)

$6 + 2k = 0 \implies k = -3$

\(P(1, m) \implies m = 4(1^{3}) - 3(1^{2}) - 6(1) + 4)

= -1

P = (1, -1)

$y = 4x^{3} + kx^{2} - 6x + 4$

$\frac{\mathrm d y}{\mathrm d x} = 12x^{2} + 2kx - 6$

At P(1, m)

$\frac{\mathrm d y}{\mathrm d x} = 12 + 2k - 6 = 0$ (parallel to the x- axis)

$6 + 2k = 0 \implies k = -3$